G4Beamline PbBi

Development of a Positron source using a PbBi converter and a Solenoid

Conclusions

1. A 0.3 (0.6) Tesla Solenoid with a diameter to allow a 9.74 (3.94) cm diameter pipe would collect a positron per thousand incident electrons on a 2mm thick LBE target with 0.25 mm thick SS windows.
2. A 15 cm long, 0.2 Tesla solenoid with a 3.94 diameter beam pipe would collect a positron per two thousand electrons impinging a 2mm thick LBE target with 0.25 mm thick SS windows.
3. A 4 Tesla Solenoid will remove beam pipe heating from scattered electrons downstream of the target when using a 3.94 cm diameter beam pipe.

Reports

Niowave_Report_11-30-2015

Task List

1.) Create a positron (10,000 positrons) and electron event file containing t,x,y,z,Px,Py,Pz for positrons exiting the solenoid and an incident Gaussian beam 1cm in diameter and with a sigma of 1cm.

compare distributions with and without solenoid.

2.) Determine the back ground when using a 3.48 diameter beam pipe and Solenoid field of 0.2 for a NaI detector placed at

3.) Experiment, install dipole and solenoid in the tunnel.

Beam Pipe Heating

PbBi_BeamPipeHeatin_2015

A 10 MeV electron beam with a radius of 0.5 cm was incident on a 2 mm thick PbBi target. The target is positioned at Z = -902 mm.

Element	dimension
Inner beam pipe radius	1.74 cm
Inner beam pipe thickness	0.165 cm
water jacket thickness	0.457 cm
outer beam pipe radius	2.362 cm
outer beam pipe thickness	0.165 cm
Solenoid inner radius	2.527 cm
Solenoid outer radius	4.406 cm

Max power deposited in beam pipe from uniform beam heating

If you assume a 1mA beam then the beam power incident on the target is

Beam Power = E(MeV) [math] \cdot [/math]I ([math]\mu[/math] A) = 10 MeV [math]\times[/math] 1000 mA = 10 kW

If the beam does not interact with the target and all the beam power is distributed uniformly along a 100 cm long beam pipe with a diameter of 3.48 cm then the power deposited per area would be

[math]10000 \mbox{W} \times \frac{1}{100 \mbox{cm}}\times \frac{1}{ \pi \times 3.48 \mbox {cm} } = 2.3 \frac{\mbox{W}}{\mbox{cm}^2}[/math]

A simulation predicts that about 8 out of 20 electrons will interact with the target and intercept a 34.8 mm diameter beam pipe surrounding the target.

[math]\mbox{P}_{\mbox{max}} = \left ( \frac{2}{5} \right ) 2.3 \frac{\mbox{W}}{\mbox{cm}^2}\lt 1 \frac{\mbox{W}}{\mbox{cm}^2}[/math]

BUT the beam does not intersect the pipe uniformly and instead can have a hot spot.

Heating along the Z-axis

GEANT4 predicts that scattered electrons, photons, and positrons (mostly scattered electrons) deposit

Energy deposited (MeV) in a 1 m long 3.48 cm diameter beam pipe surrounding a 2 mm target located at Z=-900 mm.

According to the above figure, GEANT4 predicts a total of [math]3.08\times 10^7[/math] MeV (the integral adds up the energy in each 1cm bin) of energy will be deposited in a 1m long beam pipe surrounding a 2 mm thick PbBi target located at Z=-902 mm when 20 million electrons impinge the target. The peak energy deposition is 0.3 MeV/e[math]^-[/math]

If this energy were uniformly distributed along the 5 mm thick beam pipe having a diameter of 3.48 cm then I would see

[math]3.08 \times 10^{10} \mbox{keV} \times \frac{1}{100 \mbox{cm}}\times \frac{1}{ \pi \times 3.48 \mbox {cm} } \times \frac{1}{2 \times 10^7 \mbox{e}^-}=3.5\frac{\mbox{keV}}{\mbox{cm}^2 \;\;\mbox{e}^-}[/math]

if you assume a 1 mA beam of electrons then this becomes

[math]\left ( \frac{3.5 \; \mbox{W} }{ \mbox{cm}^2 } \right) [/math]

I converted the above histogram to deposited power by 1000 mA, divide by the number of incident electrons, divide by the circumference of the beam pipe, convert the number of electrons to Coulombs, and use a unit conversion from MeV to W-s per MeV.

[math]\left(Counts \frac{\mbox{MeV}}{\mbox{cm}}\right) \times \left( \frac{1}{2 \times 10^{7} \mbox{e}^-} \right ) \times \left( \frac{1}{ \pi \times 3.48 \mbox{cm}} \right ) \times \left( \frac{1. \times 10^{-3}\mbox{ C}}{\mbox{s} }\right )\times \left( \frac{\mbox{e}^- }{1.6 \times 10^{-19}\mbox{ C}}\right ) \left( \frac{1.6 \times 10^{-13}\mbox{W} \cdot \mbox{ s}}{\mbox{MeV} }\right ) \times [/math]

If you use the above factors to weight the histogram, then the figure below shows that GEANT4 predicts a power deposition density of [math]= 4 \frac{W}{cm^2}[/math], 1 cm downstream of the target. Back scattered electrons appear to create the hottest spot of [math]= 15 \frac{W}{cm^2}[/math] about 1cm upstream of the target.

Power Deposition Zoomed in and 902 mm offset applied	Power deposition over the 1 m long beam pipe

BeamPipeHeating_4mmthick_PbBi_PositronTarget

NiowaveTargetHeatingResult_2015

Unit conversion

The energy deposited by photon, electrons, and positrons is predicted by GEANT4 and recorded in energy units of keV per incident electron on the PbBi target. To convert this deposited energy to a power you need to assume a beam current. Assuming 1 beam current of 1 mA, the conversion is given easily as

[math]\left( \frac{\mbox{keV}}{\mbox{cm}^2 \mbox{e}^-}\right) \times \left( \frac{ \mbox{e}^-}{1.6 \times 10^{-19}\mbox{C}} \right ) \times \left( \frac{1 \times 10^{-3} \mbox{C}}{\mbox{s}} \right ) \times \left( \frac{1.6 \times 10^{-16}\mbox{W} \cdot \mbox{ s}}{\mbox{keV} }\right )[/math]

[math]\left( \frac{\mbox{keV}}{\mbox{cm}^2 \mbox{e}^-}\right) = \left( \frac{\mbox{W} }{\mbox{cm}^2 } \right )[/math]

Results Table

Beam Pipe Diameter (mm)	Hot Spot ([math]MeV/e^-[/math])	Hot Spot ([math]keV/cm^2/e^-[/math])
34.8	0.35
47.5	0.24
60.2	0.20
72.9	0.16
97.4	0.12

Converter target properties

Definition of Lead Bismuth

1cm diameter target 2 mm thick PbBi

0.5 Tesla solenoid

Desire to know

Emmittance (mrad * mm)

dispersion (Delta P/P) (mradian/1000th mm/1000th)

of electrons after the PbBi target.

pole face rotation in vertical plane.

G4BeamLine and MCNPX

Solenoid

Uniform ideal Solenoid

Beam Pipe Heating with Solenoid

The energy deposited by electrons scattered into a 3.48 diameter stainless steel beam pipe (1.65 mm thick) from a PbBi target as a function of a uniform Solenoidal magnetic field.

B-field (Tesla)	Hot Spot ([math]MeV/e^-[/math])
0.0	0.35
0.3	0.35
1.0	0.35
1.5	0.22
2.0	0.10
4.0	0.002

To convert this deposited energy per incident electron on the target to a heat load in the pipe you need to divide by the area of the pipe.

A histogram is filled with 1 cm bins along the Z axis. The surface area becomes [math]1 cm \times 2 \pi 3.48/2 = 10.933 cm^2[/math]. The beam pipe diameter assumed is 3.48 cm.

When filling the histogram binned 1 cm in Z, you should weight it by the amount of depositred energy divided by the circumference of the pipe and divided by the number of incident electrons on the target (5 million). The energy units are converted to keV by multiplying the numberator by 100 or in this case dividing by 5000 instead of 5 million.

TH1F *T00N=new TH1F("T00N","T00N",100,-1000.5,-0.5)

Electrons->Draw("evt.EoutPosZ>>T00N","evt.DepE/10.088/5000")

To convert From Mev/ e- to kW/cm^2 assuming a current of 1mA (10^-3 C/s) you

[math]\left( \frac{\mbox{MeV}}{\mbox{cm}^2 \mbox{e}^-}\right) \times \left( \frac{ \mbox{e}^-}{1.6 \times 10^{-19}\mbox{C}} \right ) \times \left( \frac{1 \times 10^{-3} \mbox{C}}{\mbox{s}} \right ) \times \left( \frac{1.6 \times 10^{-13}\mbox{W} \cdot \mbox{ s}}{\mbox{MeV} }\right )[/math]

[math]\left( \frac{\mbox{keV}}{\mbox{cm}^2 \mbox{e}^-}\right) = \left( \frac{\mbox{W} }{\mbox{cm}^2 } \right )[/math]

Energy deposited (MeV) along a 1 m long beam pipe of stainless steel 1.65 mm thick.

With SS windows

Positrons->Draw("sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)","evt.BeamPosMomZ>0 && evt.BeamPosPosZ>-500 && sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)<97.4/2");

Positron Collection rates with 60 cm long Solenoid

Sample Positron Production Events

When the solenoid is 1.5 Tesla, a 10 MeV electron produces a 6.5 MeV photon that pair produces a 4.4 MeV positron and a 1 MeV electron	Same Event but this time the solenoid is 0.3 Tesla and the positron hits the beam pipe, annihilates and makes two 511 keV photons

Sample Brem event producing no positrons

When the solenoid is set to 1.5 Tesla, a 10 MeV electron produces three photons less than 1 MeV in the target, two of them compton scatter in the beam pipe	The same event but this time the electron produces only 1 photon than ionizes in the target

With SS windows

Positrons->Draw("sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)","evt.BeamPosMomZ>0 && evt.BeamPosPosZ>-500 && sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)<97.4/2");

B-field (Tesla)		34.8 mm diameter pipe	47.5	60.2	72.9	97.4
0.0	0.35	1,2,4,4,5	2,3,4,4,6	4,4,6,7,9	6,8,9,10,11	16,14,15,16,17
0.1		225,236,250,246,249=241[math] \pm[/math] 10	282,282,293,294,306=291[math] \pm[/math] 10	373,366,370,364,373=369[math] \pm[/math] 4	451,437,440,438,451=443[math] \pm[/math] 7	602,584,563,558,570=575[math] \pm[/math] 18
0.3	0.35	626,619,596,619,611 =614[math] \pm[/math] 11	720,726,706,730,717=720[math] \pm[/math] 9	871,864,840,841,834 =850[math] \pm[/math] 16	987,968,939,943,952 =958[math] \pm[/math] 20	1118,1106,1069,1067,1080=1088[math] \pm[/math] 23
0.6		929,935,949,969,961=949[math] \pm[/math] 17	1022,1031,1046,1059,1052 =1042[math] \pm[/math] 15	1120,1130,1152,1154,1146 =1140[math] \pm[/math] 15	1168,1184,1210,1221,1206 =1198[math] \pm[/math] 21	1212, 1218,1240,1254,1242=1233[math] \pm[/math] 18
1.0	0.35	1117,1085,1083,1061,1085=1086[math] \pm[/math] 20	1188,1154,1140,1111,1134=1145[math] \pm[/math] 28	1225,1190,1178,1149,1172 =1183[math] \pm[/math] 28	1243.1208,1195,1164,1184=1199[math] \pm[/math] 30	1252,1219,1206,1178,1200=1211[math] \pm[/math] 27
1.5	0.22
2.0	0.10	1198,1210,1215,1223,1176=1204[math] \pm[/math] 18	1216,1227,1235,1241,1196 =1223[math] \pm[/math] 18	1237,1243,1252,1257,1214=1241[math] \pm[/math] 17	1249,1252,1262,1266,1225 =1251[math] \pm[/math] 16	1257,1262,1270,1276,1234=1260[math] \pm[/math] 16
4.0	0.002

Positron Rates -vs- Solenoid Field for 2mm thick PbBi target and several Beam pipe diameters

Positron & Electron event files

Event files were generated assuming an ideal solenoid having an inner radius of 2.527 cm surrounding a beam pipe with a radius of 1.74 cm. Electrons impinge a 2mm thick PbBi liquid target that has a surface area of 2.54 cm x 2.54 cm. Stainless steel windows 0.25 mm thick sandwhich the PbBi target at locations Z= -90.325 and Z= -89.875 cm. The target is located at Z =-90.1 cm and the beam begins 20 cm upstream at Z = -110.1 cm. The incident electron beam is a 0.5 cm radius cylinder.

Positrons exiting the Solenoid

The solenoid design has changed such that the max field is 0.20 Tesla (0.22) and its length is 150 mm.

Apparatus

Overall Target layout	Closeup of Target Simulation Geometry. The center of the 0.25mm thick stainless steel windows are a distance of 2.25 mm from the center of the LBE target. [math]d_{USS} = d_{DSS} =[/math] 2.25 mm, [math]t_{LBE}=2[/math] mm, [math]t_{SS} =[/math] .25 mm.	An example positron production event. The yellow lines represents a sensitive detector used to record all positron events. The red line represents the incident 10 MeV electron that produces two bremsstrahlung photons shown in green. The first photon has an energy of 2.151 MeV and pair produces in the stainless steel window. The second photon has an energy of 1.393 MeV and exits the system without interacting. The first photon is created at time t= 3.323 ns. The first photon produces a 234.6 keV electron and a 894.2 keV positron at time t=3.336 ns. Only the positron has enough energy to exit the stainless steel window.

In other words I should generate position and momentum files for positrons and electrons at the Z location 15 cm downstream from the middle of the LBE target and within a 3.48 cm diameter beam pipe.

/vis/viewer/zoom 2

/gps/pos/centre 0.0 0.0 -150.

/vis/viewer/panTo -90.1 0 cm

/vis/viewer/reset

Twenty (20) million incident electrons with an energy of 10 MeV and forming a cylindrical beam with a 0.5 cm radius cylinder impinged a 2mm thick LBE target located at Z = -106 mm. The Z location of positrons exiting the beam pipe at the end of the 15 cm long solenoid is 44 mm. The positrons are 150.00 mm from the middle of the LBE target (Z=44mm).

A space delimited text file with the above events in the format of

PID, x(mm),y,z,Px,Py,Pz(MeV),t(ns)

in units of cm for distance and MeV for momentum is located at

for positrons

http://www2.cose.isu.edu/~foretony/Positrons_2mm10MeVCyl.dat

and

http://www2.cose.isu.edu/~foretony/Electrons_2mm10MeVCyl.dat

The file below contains all the positrons that were created at the target

format

A space delimited text file with the above events in the format of

Initial electron (x,y,z,Px,Py,Pz), Final electron (t,x,y,z,Px,Py,Pz), Positron location after leaving LBE target (t,x,y,z,Px,Py,Pz),Location of positron as it exits a SS window (t,x,y,z,Px,Py,Pz). Units are ns, mm, MeV/c.

http://www2.cose.isu.edu/~foretony/AllPositrons_2mm10MeVCyl.dat

Positron Distributions

Particle flight times for 20 million incident electrons on a 0.2mm thick LBE target with 0.25 mm thick stainless steel windows (no material exists after the last stainless steel window, only vacuum). The time an electron brems in the target (EscatTime) is shown in black. The time a positron is created in the target (PosTime) is shown in blue. The time a positron has traveled 15 cm traversing a uniform 0.2 Tesla Solenoidal field (PosBeamTime) is shown in fuchsia. The field exists throughout the entire world. PosBeamTimeCuts are positrons that are constrained to a beam pipe diameter of 34.8 mm. Positrons are traveling about the speed of light (30 cm/ns) so after 15 cm they arrived 0.5 nsec after the initial electron beam impinges the target.

Positrons and Electrons after the SS Exit window

The same configuration as the previous subsection except that the 1mm thick sensitive detector is placed at Z = -100.5 mm. Most of the electrons exit at Z = -100.9 mm.

A space delimited text file with the above events in the format of

PID, x(mm),y,z,Px,Py,Pz(MeV),t(ns)

in units of cm for distance and MeV for momentum is located at

for positrons

23500  Positrons in the file below

http://www2.cose.isu.edu/~foretony/PositronsAtExitWindow.dat

and

297216 electrons in the file below

http://www2.cose.isu.edu/~foretony/ElectronsAtExitWindow.dat

Solenoid Map

Inner Radiusu=

Outer Radius =

Length =

Current=

Magnetic Field Map in cylindrical coordinates (Z & R) from Niowave

Rear Window Thickness

Question: Will a thicker downstream exit window increase the positron production efficiency by providing more material for a brehm photon to pair produce in?

Positrons were counted exiting a ideal 0.2 Tesla solenoid that was 15 cm long. A ten MeV electron beam with a 0.5 cm cylindrical radius impinged a 2mm thick PbBi liquid target that had a surface area of 2.54 cm x 2.54 cm. A 0.25 mm thick stainless steel entrance window was used.

Target is at -106 mm, entrance SS window is at -108.25 mm , exit SS window is at -103.75 mm, A sensitive detector for positron is placed at Z= +44mm. The sensitive detector is a cylinder of radius 11.74 cm.

SS Exit WIndow Thickness (mm)	Positrons/Million electrons
0.0	1142,1096,1149,1073,1083 = 1109 +/- 35
0.25	774,836,800,785,798 = 798 +/- 23
0.5	693,704,713,697,715 = 704 +/- 10
1.0	587,606,548,592,550 =577 +/- 26

Conclusion 1

Positron production efficiency improves when the exit window is made thinner

Conclusion 2

You loose about 28 +/- 4 % of the positrons in the 0.25 mm thick SS exit window.

Beam Line Design

PbBi_BeamLine_Elements

goals for JLab

Positrons#Simulations